Section 2.3: Product and Quotient Rule

Objective:

Students will be able to use the product and quotient rule to take the derivative of differentiable equations

Review: Definition of Derivative

x

xfxxfxf

)()(lim)(

0x

The derivative of f at x is given by

Provided the limit exists. For all x for which this limit exists, f ’ is a function of x.

Product Rule

Theorem 2.7:The product of two differentiable function f and g is itself differentiable. Moreover, the derivative of fg is the first function times the derivative of the second , plus the second function times the derivative of the first.

You can reverse the order in which you take the derivative of the terms in the product rule.

)()()()()()( xfxgxgxfxgxfdx

d

Example #1)26)(45( 2 xxx

]45[)26(]26[)45( 22 xxdx

dxx

dx

dxx

(derivative of the second term)(first term)+(derivative of the first term)(second term)

Step 1:

)]85)(26[()]2)(45[( 2 xxxx Step 2:-Take derivative

22 16104830810 xxxxx Step 3:-Simplify

302824 2 xxStep 4:

-Simplify

Example #2

)25( 2 xxx

)5)(.25()210( )2/1(2 xxxxxStep 1:

-Take derivative

)2/1()2/3()2/3( )2/5(210 xxxx Step 2:

-Simplify

)2/1()2/3( 35.12 xx Step 3:

-Simplify

Example #3)23)(13( 3 xxx

Find the tangent line at point (-2,1) using the above equation

)2(391 xy -Plug slope & point into the point slope equationStep 6:

3)2(18)2(6)2(12 23 -plug in x=-2 from the point (-2,1) to get the slope of the

tangent line

Step 4:

)]23)(33[()]3)(13[( 23 xxxxStep 1:-Take derivative

6969393 233 xxxxxStep 2:-Simplify

318612 23 xxxStep 3:-Simplify

393362496 Step 5:-Simplify

Quotient Rule Theorem 2.8:The quotient f/g of two differentiable

functions f and g is itself differentiable at all values of x for which g(x) 0. Moreover, the derivative of f/g is given by the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator.

You can not reverse the order in which you take the derivative of the terms in the quotient rule.

≠

,)(

)()()()(

)(

)(2xg

xgxfxfxg

xg

xf

dx

d

0)( xg

Example #1

6

242

x

xy

22

22

2 )6(

]6[)24(]24[)6(]6

24[

x

xdx

dxx

dx

dx

x

x

dx

dStep 1:

(derivative of the top term)(bottom term)-(derivative of the bottom term)(top term)

(bottom term)2

22

2

)6(

)]2)(24[()]4)(6[(

x

xxxStep 2: -Take derivative

3612

244424

2

xx

xxStep 3: -Simplify

Example #2

7

)/1(53

x

xy

)7(

)]/1(5[3

xx

xxStep 1: -Get rid of fraction in the numerator by

multiply the numerator and denominator by x

xx

x

7

154

Step 2: -Simplify

24

34

)7(

)74)(15()5)(7(

xx

xxxx

Step 3: -Take derivative

258

34

4935

7415

xxx

xx

Step 4: -Simplify

Example #38

64

2

x

xy

Find tangent equation at point (-1,3)

24

234

)8(

)6(4)8(2

x

xxxxStep 1: -Take derivative

6416

2441628

355

xx

xxxxStep 2: -Simplify

6416

1624248

35

xx

xxxStep 3: -Simplify

81

10

64)1(16)1(

)1(16)1(24)1(248

35

Step 4: -plug in x=1 from the point (-1,3) to find the slope of the tangent line

)1(81

103 xyStep 5: -plug the slope and point into the point

slope formula

Combining the Product Rule & Quotient Rule

9

)84)(23( 2

x

xx

8118

1682412)9)(16242412(2

2322

xx

xxxxxxxStep 2:

-Simplify

8118

1682412144216216108162424122

2322233

xx

xxxxxxxxxxStep 3:

-Simplify

8118

200144332242

23

xx

xxxStep 4:

-Simplify

*For this type of problem use the quotient rule and with in the quotient rule use the product rule to take the derivative of the numerator

2

22

)9(

)84)(23()1()9()23(8)84(3

x

xxxxxxStep 1:

-Take derivative

Product rule for derivative of the numerator